A body thrown vertically up to reach its maximum height in t second. The total time from the time of projection to reach a point at half of its maximum height while returning (in second) is

Question

A body thrown vertically up to reach its maximum height in t second. The total time from the time of projection to reach a point at half of its maximum height while returning (in second) is (A) √2 t  (B)  1 + (1/√2) t  (C) (3t/2)  (D) (t/√2)

Tardigrade Admin · Accepted Answer

The ball is thrown vertically upwards then according to equation of motion

(0)^{2} - u^{2} = - 2 g h

...(i)
and

0 = u - g t

....(ii)
From Eqs. (i) and (ii),

h = \frac{g t ^{2}}{2}

When the ball is falling downwards after reaching the maximum height

s = u t^{'} + \frac{1}{2} g (t^{'})^{2}

\frac{h}{2} = (0) t^{'} + \frac{1}{2} g (t^{'})^{2}

\Rightarrow t^{'} = \frac{h}{g}

t^{'} = \frac{t}{2}

Hence, the total time from the time of projection to reach a point at half of its maximum height while returning =

t + t^{'}

= t + \frac{t}{2}

Q. A body thrown vertically up to reach its maximum height in $t$ second. The total time from the time of projection to reach a point at half of its maximum height while returning (in second) is

$2 t$

$1 + \frac{1}{2} t$

$\frac{3 t}{2}$

$\frac{t}{2}$

Q. A body thrown vertically up to reach its maximum height in t second. The total time from the time of projection to reach a point at half of its maximum height while returning (in second) is

2​t

1+2​1​t

23t​

2​t​

Q. A body thrown vertically up to reach its maximum height in $t$ second. The total time from the time of projection to reach a point at half of its maximum height while returning (in second) is

$2 t$

$1 + \frac{1}{2} t$

$\frac{3 t}{2}$

$\frac{t}{2}$